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I am interested in the asymptotics of $\newcommand{agm}{\operatorname{agm}}\agm(1,x)$ as $x\to\infty$, where $\agm$ is the arithmetic geometric mean. I know that

$$\sqrt x\le\agm(1,x)\le\frac{1+x}2$$

Thus, $\agm(1,x)\in\mathcal O(x)$. I then noticed that

$$\agm(1,x)=\agm\left(\sqrt x,\frac{1+x}2\right)=\sqrt x\agm\left(1,\frac{\sqrt x+\frac1{\sqrt x}}2\right)$$

For large $x$, I imagine that we have

$$\agm\left(1,\frac{\sqrt x+\frac1{\sqrt x}}2\right)\sim_\infty\agm\left(1,\frac{\sqrt x}2\right)$$

And if $\agm(1,x)\sim_\infty\alpha x^\epsilon$, then

$$x^\epsilon=x^{\frac12(\epsilon+1)}\\\epsilon=\frac12(\epsilon+1)\\\epsilon=1$$

Is this correct? And if so, how do I calculate $$\alpha=\lim_{x\to\infty}\frac{\agm(1,x)}x\ ?$$

It certainly appears to be the case that $\alpha<1$, though I cannot conclude much more than that.

Perhaps one might find the integral form to be useful:

$$\agm(1,x)=\frac\pi{2I(1-x^2)}$$

where

$$I(x)=\int_0^{\pi/2}\frac{dt}{\sqrt{1-x\sin^2(t)}}=\int_0^1\frac{dt}{\sqrt{(1-t^2)(1-xt^2)}}$$

Martin Sleziak
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Simply Beautiful Art
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  • Looks like you are going after this agm in a big way. +1 – Paramanand Singh Aug 19 '17 at 09:34
  • Modified incorrect statement about $\text{agm}(1,x)$ being in $O(x^\epsilon)$ for $\epsilon\in[0.5,1]$ (except if $\epsilon=1$). Unrelatedly, I have qualms about the fact that you use \newcommand in the title, but, not knowing the exact dogma about this, I did not touch it. – Did Aug 19 '17 at 10:23
  • @Did Uh, but I wanted that part. The bounds $\sqrt x$ and $\frac{1+x}2$ meant that $\agm(1,x)\in\mathcal O(x^\epsilon)$ for some $\epsilon\in[0.5,1]$, right? – Simply Beautiful Art Aug 19 '17 at 11:24
  • Actually the first time I've used newcommand, so I don't know much about it either. – Simply Beautiful Art Aug 19 '17 at 11:24
  • The bounds $\sqrt{x}\leqslant u(x)\leqslant\frac12(1+x)$ for every $x$ (or only for every $x$ large enough) mean that $$u(x)=O(x)\qquad u(x)=\Theta(\sqrt{x})$$ but not the odd formulation you suggest. The only properties $$u(x)=O(x^\epsilon)$$ that these bounds ensure are those such that $$\epsilon\geqslant1$$ – Did Aug 19 '17 at 11:26
  • @Did Hm, I had thought that it would translate into $\exists\epsilon\in[0.5,1]\text{ s.t. }\agm(1,x)\in\mathcal O(x^\epsilon)$, namely $\epsilon=1$. – Simply Beautiful Art Aug 19 '17 at 11:30
  • Well, perhaps I should go back and check the definitions of these things. Thanks. – Simply Beautiful Art Aug 19 '17 at 11:32

2 Answers2

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For every $x>1$, $$\mathrm{agm}(1,x)=x\cdot\mathrm{agm}(1,x^{-1})=\frac{\pi x}{2K(u(x))}$$ where $$u(x)^2=1-x^{-2}$$ and $K$ denotes the complete elliptic integral of the first kind.

When $x\to\infty$, $u(x)\to1$. The asymptotic expansion of $K(k)$ when $k\to1$ reads $$K(k)=-\frac12\log|1-k|+O(1)$$ hence, when $x\to\infty$, $$\mathrm{agm}(1,x)=\frac{\pi x}{-\log|1-u(x)|+O(1)}=\frac{\pi x}{2\log x+O(1)}$$ in particular,

$$\lim_{x\to\infty}\frac{\log x}x\cdot\mathrm{agm}(1,x)=\frac\pi2$$

Did
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2

The fundamental asymptotic here is $$\frac{\pi} {2\operatorname {agm} (1,k)}=K(k') = \log\frac{4}{k}+o(1)$$ as $k\to 0^{+}$. This can be written as $$\frac{\pi} {2k\operatorname {agm} (1,1/k)}=\log\frac{4}{k}+o(1)$$ Putting $x=1/k$ we see that $$\frac{\pi x} {2\operatorname {agm} (1,x)}=\log 4x+o(1)$$ as $x\to\infty $. This implies that your function $\operatorname {agm}(1,x)$ behaves like $\pi x/\log 16x^{2}$ as $x\to\infty$.

Another way to look at this is in terms of nome $q=e^{-\pi K'/K} $ and Jacobi's theta functions. Note that $$k=\frac{\vartheta_{2}^{2}(q)}{\vartheta_{3}^{2}(q)}=\frac{4\sqrt{q}\psi^{2}(q^{2})}{2K/\pi}$$ where $\psi$ is one of Ramanujan's theta functions. When $q\to 0^{+}$ then $2K/\pi\to 1,\psi(q^{2})\to 1$ so that $k/4\sqrt{q}\to 1$. And since $\pi K'/K=\log q^{-1}$ it follows that $K'/\log \sqrt{q} \to - 1$. And this combined with $k/4\sqrt{q}\to 1$ gives us $K'/\log(4/k)\to 1$ as $k\to 0$.

You may also obtain the asymptotic for $K'$ by noting that both $K(k) $ and $K(k') $ satisfy the same differential equation and an analysis of indicial equation gives the link between $K$ and $K'$ namely $$K(k') =\frac{2K(k)} {\pi}\log\frac{4}{k} -2\left\{\left(\frac{1}{2}\right)^{2}\left(\frac{1}{1\cdot 2}\right)k^{2}+\left(\frac{1\cdot 3}{2\cdot 4}\right)^{2}\left(\frac{1}{1\cdot 2}+\frac{1}{3\cdot 4}\right)k^{4}+\dots\right\}$$

It is of interest to note that Ramanujan presented the above asymptotic in the following grand form $$\lim_{x\to 0^{+}}\frac{1}{x}\exp\left(-\pi\cdot\dfrac{{}_{2}F_{1}\left(\dfrac{1}{2},\dfrac{1}{2};1;1-x\right)}{{}_{2}F_{1}\left(\dfrac{1}{2},\dfrac {1}{2};1;x\right)}\right)=\frac{1}{16}$$ and proved it using certain identities relating hypergeometric functions. Using this formula and some more hypergeometric identities (mainly Gauss-Landen quadratic transformation) Ramanujan inverted the relation $$q=\exp\left(-\pi\cdot\dfrac{{}_{2}F_{1}\left(\dfrac{1}{2},\dfrac{1}{2};1;1-x\right)}{{}_{2}F_{1}\left(\dfrac{1}{2},\dfrac {1}{2};1;x\right)}\right)$$ to get $$x=\frac{\vartheta_{2}^{4}(q)} {\vartheta_{3}^{4}(q)}, \, {}_{2}F_{1}\left(\frac{1}{2},\frac {1}{2};1;x\right)=\vartheta_{3}^{2}(q)$$ and further deduced the transformation formula for theta functions. This is quite unlike the modern approaches (based on Poisson summation formula) and Jacobi's approach (based on integral transformations). Ramanujan's approach is more powerful and it led him to consider expressions $$q_{r}(x) = \exp\left(-\frac{\pi} {\sin(\pi/r) } \cdot\dfrac{{}_{2}F_{1}\left(\dfrac{1}{r},\dfrac{r-1}{r};1;1-x\right)}{{}_{2}F_{1}\left(\dfrac{1}{r},\dfrac {r-1}{r};1;x\right)}\right)$$ for $r=3,4,6$ and develop alternative theories of theta functions.

Paramanand Singh
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